The one at the base would be much older due to the law of super position, and the rock at the top would be much newer,again, due to the law of super position.
Answer:
Wavelength of the sound wave that reaches your ear is 1.15 m
Explanation:
The speed of the wave in string is

where T= 200 N is tension in the string ,
=1.0 g/m is the linear mass density


Wavelength of the wave in the string is

The frequency is

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)

Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
Answer:
nucleus is the center of atom
Answer:
The width of the central bright fringe is 7.24 mm.
Explanation:
Given that,
Wavelength = 632.8 nm
Width d= 0.350 mm
Distance between screen and slit D= 2.00 m
We need to calculate the distance
Using formula of distance

Put the value into the formula


We need to calculate the width of the central bright fringe
Using formula of width

Put the value into the formula


Hence, The width of the central bright fringe is 7.24 mm.